2x^2+8x-50=0

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Solution for 2x^2+8x-50=0 equation:



2x^2+8x-50=0
a = 2; b = 8; c = -50;
Δ = b2-4ac
Δ = 82-4·2·(-50)
Δ = 464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{29}}{2*2}=\frac{-8-4\sqrt{29}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{29}}{2*2}=\frac{-8+4\sqrt{29}}{4} $

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